Class Solution

  • All Implemented Interfaces:

    
    public final class Solution
    
                        

    1889 - Minimum Space Wasted From Packaging\.

    Hard

    You have n packages that you are trying to place in boxes, one package in each box. There are m suppliers that each produce boxes of different sizes (with infinite supply). A package can be placed in a box if the size of the package is less than or equal to the size of the box.

    The package sizes are given as an integer array packages, where packages[i] is the size of the <code>i<sup>th</sup></code> package. The suppliers are given as a 2D integer array boxes, where boxes[j] is an array of box sizes that the <code>j<sup>th</sup></code> supplier produces.

    You want to choose a single supplier and use boxes from them such that the total wasted space is minimized. For each package in a box, we define the space wasted to be size of the box - size of the package. The total wasted space is the sum of the space wasted in all the boxes.

    • For example, if you have to fit packages with sizes [2,3,5] and the supplier offers boxes of sizes [4,8], you can fit the packages of size-2 and size-3 into two boxes of size-4 and the package with size-5 into a box of size-8. This would result in a waste of (4-2) + (4-3) + (8-5) = 6.

    Return the minimum total wasted space by choosing the box supplier optimally , or -1 if it is impossible to fit all the packages inside boxes. Since the answer may be large , return it modulo <code>10<sup>9</sup> + 7</code>.

    Example 1:

    Input: packages = 2,3,5, boxes = \[\[4,8],2,8]

    Output: 6

    Explanation: It is optimal to choose the first supplier, using two size-4 boxes and one size-8 box.

    The total waste is (4-2) + (4-3) + (8-5) = 6.

    Example 2:

    Input: packages = 2,3,5, boxes = \[\[1,4],2,3,3,4]

    Output: -1

    Explanation: There is no box that the package of size 5 can fit in.

    Example 3:

    Input: packages = 3,5,8,10,11,12, boxes = \[\[12],11,9,10,5,14]

    Output: 9

    Explanation: It is optimal to choose the third supplier, using two size-5 boxes, two size-10 boxes, and two size-14 boxes.

    The total waste is (5-3) + (5-5) + (10-8) + (10-10) + (14-11) + (14-12) = 9.

    Constraints:

    • n == packages.length

    • m == boxes.length

    • <code>1 <= n <= 10<sup>5</sup></code>

    • <code>1 <= m <= 10<sup>5</sup></code>

    • <code>1 <= packagesi<= 10<sup>5</sup></code>

    • <code>1 <= boxesj.length <= 10<sup>5</sup></code>

    • <code>1 <= boxesk<= 10<sup>5</sup></code>

    • <code>sum(boxesj.length) <= 10<sup>5</sup></code>

    • The elements in boxes[j] are distinct.

    • Nested Class Summary

      Nested Classes 
      Modifier and Type Class Description
    • Field Summary

      Fields 
      Modifier and Type Field Description
    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()
    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description
    • Method Summary

      Modifier and Type Method Description
      final Integer minWastedSpace(IntArray packages, Array<IntArray> boxes)
      • Methods inherited from class java.lang.Object

        clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait